[英]Typescript return type inference
In the following code snippet, typescript is able to infer the return type correctly: 在以下代码片段中,typescript能够正确推断返回类型:
{
type IOFunction< OUTPUT> = (input: number) => OUTPUT
function createIOFunction<OUTPUT>(input: number, converter: IOFunction<OUTPUT>): OUTPUT {
return converter(input)
}
const x = createIOFunction(12, num => num + '!')
// x is of type 'string'. Typescript was able to infer the OUTPUT type correctly AND without me specifying what the output type should be
const y = createIOFunction(24, num => num * 2)
// y is of type 'number'
}
How can I achieve this with following construct? 如何通过以下构造实现此目的?
{
type IOFunction<INPUT> = (input: INPUT) => any /* <-- What goes here ?? */
// I don't want the type to be IOFunction<INPUT, OUTPUT> = (input: INPUT) => OUTPUT
const convert: IOFunction<number> = num => num + '!'
const x = convert(12)
// x is of type any, not string
// how can x be inferred?
}
Here a more complex example (as requested): 这是一个更复杂的例子(根据要求):
interface STATE {
value: number
}
const myState: STATE = {
value: 12
}
type FAC_FUNCTION<S> = (state: S) => any
const factory: FAC_FUNCTION<STATE> = state => (
{
hello() {
return 'Hello ' + state.value;
}
})
const toolbox = factory(myState)
// toolbox.hello() <!--- not typed :(
type FACTORY_FUNCTION<S, OUTPUT> = (state:S) => OUTPUT
function bindFactory<S, OUTPUT>(state:S, fac: FACTORY_FUNCTION<S, OUTPUT>) {
return fac(state)
}
const toolbox2 = bindFactory(myState, state => ({
hello() {
return 'Hello ' + state.value
}
}))
toolbox2.hello() // typed :)
The toolbox is not typed, toolbox2 is. 工具箱未输入,toolbox2是。 I want to bind the state to the specific functions.
我想将状态绑定到特定的函数。 In the end i wan't the user to write something like
const toolbox = bindFactory(state, factory)
. 最后,我不想让用户编写像
const toolbox = bindFactory(state, factory)
。
Sorry, but this is the best complex example I came up with 对不起,这是我提出的最好的复杂例子
Ok, i figured a way - and it is actually very simple. 好吧,我想办法 - 实际上非常简单。 All you have to provide are the type definitions of the input parameters and let the compiler figure out the rest.
您需要提供的只是输入参数的类型定义,让编译器弄清楚其余部分。
So instead of 而不是
type FAC_FUNCTION<S> = (state: S) => any
const factory: FAC_FUNCTION<STATE> = state => (
{
hello() {
return 'Hello ' + state.value;
}
})
const toolbox = factory(myState)
use: 采用:
const factory = (state:PROPS) => (
{
hello() {
return 'Hello ' + state.value;
}
})
const toolbox = factory(myState)
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.